CBSE Class 12 Chemistry Basic Concepts of Chemistry Revision Cards

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Q2. Fill in the blanks :

a) No. of significant figure in 0.0025 is ------------

Basic Concepts in Chemistry – Revision Cards

Card 1

(a) Match the following.

Answer: 1 – e, 2 – a, 3 – d, 4 – b, 5 – c.

(b) Match the following.

Answer: 1 – b, 2 – c, 3 – d, 4 – a.

Card 2

Fill in the blanks.

1. The number of significant figures in 0.0025 is ______.
2. The value of 20.058 rounded up to 4 significant figures is ______.
3. The sum of the mole fractions of all the components present in a solution is ______.
4. The unit of molarity is ______.
5. Out of molarity and molality, ______ changes with temperature.
6. The mass of one mole of a substance in grams is called ______.
7. One mole of any gas at STP occupies ______.
8. The simplest whole-number ratio of the various atoms present in one molecule of a compound is called ______.
9. The number of atoms present in 5 u of He is ______.
10. The number of moles of \(\mathrm{CO_2}\) present in 4.4 g of \(\mathrm{CO_2}\) is ______.

Answer:

1. 2 (the zeros before the 2 are not significant).
2. 20.06
3. 1
4. mol \(\mathrm{L^{-1}}\) (also written M).
5. molarity (the volume of the solution changes with temperature).
6. molar mass.
7. 22.4 L (at 273 K and 1 atm; it is 22.7 L at 273.15 K and 1 bar).
8. the empirical formula.
9. 1.25 (one He atom has a mass of 4 u, so 5 u / 4 u = 1.25).
10. 0.1 mol (4.4 g / 44 g \(\mathrm{mol^{-1}}\)).

Card 3

Answer the following in brief.

In 3 moles of ethane, calculate the number of moles of carbon atoms.

Answer: Ethane is \(\mathrm{C_2H_6}\), so 1 mol has 2 mol of C atoms. Hence 3 mol of ethane has 6 mol of carbon atoms.

What is the mass of one amu in grams?

Answer: 1 amu = \(1.66\times10^{-24}\) g.

What is the SI unit of pressure?

Answer: The pascal (Pa), which is \(\mathrm{N\,m^{-2}}\).

What is the empirical formula of benzene?

Answer: Benzene is \(\mathrm{C_6H_6}\), so the empirical formula is CH.

What is the equivalent mass of sulphuric acid?

Answer: Equivalent mass = molar mass / basicity = 98 / 2 = 49 g \(\mathrm{eq^{-1}}\).

One gram of hydrogen reacts with 9 g of oxygen to form water. Which one is the limiting reagent?

Answer: The equation is \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow 2\mathrm{H_2O}\). 1 g of hydrogen needs 8 g of oxygen, but 9 g of oxygen is present. So hydrogen is the limiting reagent (and 9 g of water is formed).

What is the mass of one molecule of \(\mathrm{CO_2}\)?

Answer: Mass of one molecule = 44 g / \(6.022\times10^{23}\) = \(7.31\times10^{-23}\) g.

What is the mass of one atom of sodium?

Answer: Mass of one atom = 23 g / \(6.022\times10^{23}\) = \(3.82\times10^{-23}\) g.

What is the relation between molar mass and vapour density?

Answer: Molar mass = 2 × vapour density.

10 volumes of hydrogen gas react with 5 volumes of oxygen gas. How many volumes of water vapour will be produced?

Answer: By Gay-Lussac's law, \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow 2\mathrm{H_2O}\), so 10 volumes of hydrogen and 5 volumes of oxygen give 10 volumes of water vapour.

Card 4

A solution is prepared by adding 2 g of a substance to 18 g of water. Calculate the mass percentage of the solute.

Answer: \(\text{Mass \% }=\dfrac{2}{2+18}\times100=10\%\). The mass percentage of the solute is 10%.

Calculate the molarity of NaOH in the solution prepared by dissolving 4 g of it in enough water to form 250 mL of the solution.

Answer: Moles of NaOH = 4 g / 40 g \(\mathrm{mol^{-1}}\) = 0.1 mol.
\(M=\dfrac{0.1\ \mathrm{mol}}{0.25\ \mathrm{L}}=0.4\ \mathrm{M}\)

Calculate the number of moles, molecules and atoms present in 62 g of phosphorus (\(\mathrm{P_4}\)).

Answer: Molar mass of \(\mathrm{P_4}\) = 124 g \(\mathrm{mol^{-1}}\).
Moles = 62 / 124 = 0.5 mol.
Molecules = \(0.5\times6.022\times10^{23}=3.011\times10^{23}\).
Atoms = 4 × \(3.011\times10^{23}=1.2044\times10^{24}\).

The density of a 3 M solution of NaCl is 1.25 g \(\mathrm{mL^{-1}}\). Calculate the molality of the solution.

Answer: Take 1 L of solution. Mass of the solution = 1000 × 1.25 = 1250 g. Mass of NaCl = 3 × 58.5 = 175.5 g. Mass of water = 1250 − 175.5 = 1074.5 g = 1.0745 kg.
Molality = 3 mol / 1.0745 kg = 2.79 m.

Calculate the amount of carbon dioxide that could be produced when
(a) 1 mole of carbon is burnt in air,
(b) 1 mole of carbon is burnt in 16 g of dioxygen,
(c) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer: The reaction is \(\mathrm{C}+\mathrm{O_2}\rightarrow\mathrm{CO_2}\).
(a) Oxygen in air is in excess, so 1 mol of carbon gives 1 mol of \(\mathrm{CO_2}\) = 44 g.
(b) 16 g of \(\mathrm{O_2}\) is 0.5 mol, and it is the limiting reagent. So 0.5 mol of \(\mathrm{CO_2}\) = 22 g is produced.
(c) Again 16 g (0.5 mol) of \(\mathrm{O_2}\) is the limiting reagent. So 0.5 mol of \(\mathrm{CO_2}\) = 22 g is produced.

How are 0.50 mole \(\mathrm{Na_2CO_3}\) and 0.50 M \(\mathrm{Na_2CO_3}\) different?

Answer: 0.50 mole of \(\mathrm{Na_2CO_3}\) is an amount of substance: 0.50 × 106 = 53 g of it. 0.50 M \(\mathrm{Na_2CO_3}\) is a concentration: 0.50 mol (53 g) of it is present in every litre of the solution.

Find the amount of oxalic acid required to prepare 100 mL of M/20 oxalic acid solution.

Answer: M/20 = 0.05 M. Moles needed = 0.05 × 0.1 L = 0.005 mol. Crystalline oxalic acid, \(\mathrm{H_2C_2O_4\cdot2H_2O}\), has a molar mass of 126 g \(\mathrm{mol^{-1}}\). Mass = 0.005 × 126 = 0.63 g. (If anhydrous oxalic acid, 90 g \(\mathrm{mol^{-1}}\), is used, the mass is 0.45 g.)

Card 5

State whether the following are true or false.

1. An element contains particles of only one type, which may be atoms or molecules.
2. 1 mol of any substance is equal to \(6.022\times10^{23}\) of the respective particles or entities.
3. The number of moles of a solute present in one litre of the solution is called molality.
4. Molarity is a better term to express the concentration of a solution.
5. A 1 M aqueous solution of a solute is more concentrated than a 1 m solution of the same solute.

Answer:

1. True.
2. True.
3. False. It is called molarity. Molality is the number of moles of solute per kg of solvent.
4. False. Molality is the better term, because it does not change with temperature (mass does not change, but volume does).
5. True. In 1 M, 1 mol of solute is in one litre of solution, so the solvent is less than 1 kg. So the solute per kg of solvent is more than in a 1 m solution.

3. The no. of moles of any solute present in one liter of solution is called molality.(T/F)